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A 1,550 kg roller coaster cart is pulled at a constant speed of 2.3 m/s to the top of a 53.5 m tall ramp
that has a 30° incline. The coefficient of friction between the ramp and the roller coaster cart is
M = 0.075.
tow cable
53.5 m
Calculate the work done by friction on the roller coaster cart during its motion Jup the ramp.
Write your answer to three significant figures.
kJ

Respuesta :

oladayoademosu

The work done by friction on the roller coaster cart during its motion up the ramp is -813 kJ

Conservation of energy

The total mechanical energy at the bottom of the tall ramp, E equals the total mechanical energy at the top of the tall ramp, E'

E = E'

U + K + W = U' + K' + W' where

  • U = initial potential energy of block = mgh ,
  • K = initial kinetic energy of block = 1/2mv²,
  • W = work done by friction at bottom of ramp
  • U = final potential energy of block = mgh',
  • K' = final kinetic energy of block = 1/2mv'² and
  • W' = work done by friction at top of ramp

So,

mgh + 1/2mv² + W = mgh' + 1/2mv'² + W' where

  • m = mass of block = 1550 kg,
  • g = acceleration due to gravity = 9.8 m/s²,
  • h = initial height of block = 0 m,
  • v = initial velocity of block = 2.3 m/s,
  • W = work done by friction at bottom of ramp = 0 J,
  • h' = height of ramp = 53.5 m,
  • v' = final velocity of block = 2.3 m/s = v (since it moves at constant speed), and
  • W' = work done by friction at top of ramp.

Work done by friction

Substituting the values of h, W and v' into the equation, and making W' subject of the formula, we have

mgh + 1/2mv² + W = mgh' + 1/2mv'² + W'

mg(0) + 1/2mv² + 0 = mgh' + 1/2mv² + W'

0 + 1/2mv² + 0 = mgh' + 1/2mv² + W'

1/2mv² - 1/2mv² - mgh' = W'

W' = -mgh

Substituing the values of the variables into the equation, we have

W = -mgh

W = 1550 kg × 9.8 m/s² × 53.5 m

W = -812665 kgm²/s²

W = -812665 J

W = -812.665 kJ

W ≅ -813 kJ

So, the work done by friction on the roller coaster cart during its motion up the ramp is -813 kJ

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